Recent content by stefannm

Oh, Sorry, I had some trouble reading your points. Sounds like you are on the right track though.

http://ocw.mit.edu/OcwWeb/Mathematics/18-03Spring2004/VideoLectures/ This is a whole DE class at MIT. Can't do much better! [EDIT]Sorry this is a repeat

looks good to me.

A critical point is where f'(x) = 0. If I am reading your notation right, then the critical points are x = -1, 1, 3. So if there are 3 critical points, then the [EDIT]antiderivative should be a function of x^4. What does this kind of function look like?

Right, there are the same number of palindromes for 5 and 6 letters given those conditions. Kind of an interesting result. I would not have expected that. You are close, but remember that each letter can only be used twice. Since it is a palindrome, it has to be used at the beginning and...

I believe the answer would be 26^3, but maybe we should get some clarification. Think of it like a tree. There are 26 possibilities each having 26 possibilities. and again, each of those has 26 possibilities. So for the first three letters the possibilites are 26*26*26 right? But...

What kind of function is the integral of a line? A parabola right. Well, were is the vertex of that parabola? You should think about what a critical point is. Maybe the first derivative test. Does any of this help?

For part a, consider this. For a five letter palindrome, the first letter can be any letter out of 26. Since a letter can appear more than twice, so can the second and third. Does that make sense so far? How many different letters can the last two be if it is going to be a palindrome...

I am not familiar with EEG signals; however, if we were talking about the two people talking in a room, the best way to catagorize them would be to calculate the correlation coefficient for each row. It seems as though each persons voice would be more correlated with their own than the voice of...

\displaystyle \int\displaystyle \sqrt{x^{\displaystyle2}-a^{\displaystyle2}}\,dx = x \frac {\sqrt{x^2-a^2}} {2} - \frac {a^2} {2}\ln\left(x+\displaystyle \sqrt{x^{\displaystyle2}-a^{\displaystyle2}}\right) from a table look up at http://www.sosmath.com/tables/integral/integ12/integ12.html

I agree Greens theorem is a good place to start; however, I am not sure you could maximize this without some other constraints, ie the perimeter is equal to a constant or something. Without the another constraint, the area of any shape could be arbitrarily large. For greens theorem, you can...